Problem 1175

I. Solution: Subtract the difference of the sums of the groups to be formed, 9, from the sum of the given numbers, 86. Dividing the remainder, 77, into two equal parts, we get the sum of the smaller group. But we immediately see that the desired distribution is impossible, because 77 is odd, its half is not an integer, and all the given numbers are integers.

Remarks. 1. Essentially, those who proved the impossibility of the desired distribution by denoting the sum of the numbers in the smaller group by $x$ - thus the sum of the other group is $x+9$, and from the equation $x+(x+9)=86$ showed that its solution is a fraction.

2. Several contestants thought they solved the problem by omitting the 9 among the given numbers, and showed that the remaining numbers cannot be divided into two groups with equal sums. However similar this idea may be to the above solution, - it is not identical to it, and it is not a complete proof. From this, we only see that there is no solution to the problem in which the 9 is part of the larger sum group. However, a solution is not impossible in which the 9 is a member of the smaller sum group. For example, if instead of the given numbers, we need to divide the numbers $9,2,2,2,2,2,2,2,2$ into two groups under the same condition, this is possible: the smaller sum group contains only the 9, and the other group contains the 2s, although - as is easy to see - the 9 cannot be a member of the larger sum group here either.
3. Generally: it is not possible to divide given integers into two groups such that the difference of the sums of the groups is a given integer, if subtracting the required difference from the sum of the numbers results in an odd number. But if an even number is obtained this way, it does not follow that the desired grouping is possible. For example, the set of numbers $9,2,2,2,2,2$ cannot be divided into two groups according to our condition, although the difference between their sum and the required difference is divisible by 2: $(19-9): 2=5$.
4. Our finding can be generalized to more groups as follows. A necessary - but, as we have seen, not sufficient - condition for the divisibility of given integers into $k$ groups with given integer differences is that subtracting the sum of the differences from the sum of the numbers results in a number divisible by $k$. (This sum is to be understood as the differences of the sums of all groups from the sum of a certain, designated group of numbers, including their signs.)

II. Solution: The problem requires us to form two groups from the listed integers such that the difference of the sums of the numbers in them is 9, and the two sums together add up to 86. This is impossible because if the difference of two integers is odd, then the two numbers are of opposite parity, and therefore their sum is also odd.